Concentration Problem (with my solution, tell me if im right)?

Question

An excess of Calcium nitrate (NO3) was added to 50.0 mL of sodium fluoride, NaF, solution. The mass of the calcium fluoride precipitate, CaF2, produced was 2.93 g. What is the concentration of the Sodium Fluoride solution?
Ca(NO3)(aq) + 2 NaF(aq) ---> CaF2(s) + 2 NaNO3(aq)

what i thought of doing was... C = n/v

n = 2.93 / the molar mass of NaF
v = 50.0mL -> 0.05mL

n = 2.93 / 33 = 0.8
V = 0.05

n/v = 1.7

Answer 1

I am afraid you are wrong.
From the looks of it tou want a simple solution to the problem so we consider that precipitation is 100% (otherwise you need the Ksp of CaF2)

First you have to find how many moles of CaF2 are formed:
n=m/MW=2.93/78.08=0.0375 moles CaF2

From the stoichiometry of the reaction you know that
2 moles of NaF give 1mole of CaF2, thus
x mole of NaF give 0.0375 moles of CaF2

thus x=2*0.0375=0.075 moles of NaF

Therefore C=n/v=0.075/0.05=1.5 M NaF

You should be careful and do the analogy with moles not grams

Answer 2

concentrated of NaF solution is 17.06%